Question

(x Carry out a Breusch-Pagan LM test for heteroskedasticity for the model: β8male + 11 Briefly explain how the test works and the implications ofyou「result. [Hint: see rea uhatsa totwrk educ age agesa spwrk75 gdhlth ynakid male 706 1.62 SourceI df Number of obs MS 8 , 697 ) 8 2.0691e+11 Prob >F Model 1.6553e+12 0.1163 8.9178e+13 0.0182 Residual 697 1.2794e+11 R-squared Adj R-squared 0.0070 + = Total 0833e+13 705 1.2884e+11 Root MSE 3.6e+05 Coef. uhatsaI Std. Err. [95% Conf. Interval] totwnk 1 17.09085 15.61925 1.09 0.274 -13.57557 47.75727 educ-9030.0515165.27 age -8175.251 979 agesa59.24164116.5895 spwrk75 ..-.29976.39 27630.83-1.08 0.278 -84226.0324273.26 gdhlth 21977.36 45480.56 1.750.081 19171.4 1111.302 11063 -27413.5 9798.57 0.83 0.404 0.51 0.612 -169.6671 288.1504 111272.7 67317.96 -0.48 0.629 yngkid 1702.65643038.76 male I -41564.2329890.5 86203.81 0.04 0.968 -82798.5 1.39 0.165 -100250.4 17121.97 cons529497.8 205353 2.58 0.010 126313.1 932682.4

Answer 1

Breusch Pagan test statistic = nR^2 = 706 * 0.0182 = 12.85

This follows a chi square statistic with p-1 degrees of freedom where p is the number of parameters.

p = 9

Chi square p value at 8 degrees of freedom for 12.85 = 11.7%. So we fail to reject the null hypothesis that there is no heteroskedasticity, ie it is homoskedastic.

Breusch pagan test carries out a regression of residuals of the original model against the independent variables. If e be the residuals of the regression of y (dependent) with x (independent) then, Breusch pagan test establishes the relation;

$e^{2} = f(x) + u = \theta_{0} + \theta_{1}x_{1} + \theta_{2}x_{2}+...+u$

The Breusch Pagan test statistic (nR^2) is a chi square test statistic with p-1 degress of freedom, where p is the number of parameters in the model.

The null hypothesis is that there is no heteroskedasticity in the model.